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Hydraulics / fluid dynamics

  • Thread starter fayan77
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Homework Statement


Screen Shot 2018-10-04 at 7.43.51 PM.png


Homework Equations


y1+P1/##\gamma##+##\alpha##v12/2g = y2+P2/##\gamma##+##\alpha##v22/2g + hL

The Attempt at a Solution


so it gives me pressure at p2 but not p1. I tried using Bernoulli's equation to find p1 i did plugged into energy equation above and got an answer but it did not make sense i got hL = .01 ft (I assumed that v1 = v2 but i do not think that is valid)
 

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  • #2
haruspex
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got an answer
Please post your working (and not as an image - those are for textbook extracts and diagrams).
 
  • #3
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...That is the problem exercise. Do you want me to type it instead?
I would like some direction so I can try it and post my work. Am I correct to use Bernoulli's equation to find P1? and will velocities be the same in both sections?
 
  • #4
haruspex
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That is the problem exercise. Do you want me to type it instead?
The image you posted is fine. I am asking you to post your own working, but not as an image. I find that if I do not say that up front the working gets posted as an image which is hard to read and even harder to comment on.
Am I correct to use Bernoulli's equation to find P1? and will velocities be the same in both sections?
Yes and yes.
 
  • #5
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ok, So I start off with Bernoulli's equation solving for P1. Because v1 = v2 the kinetic energy term will cancel out leaving us with

P1 + ##\rho##gh1 = P2 + ##\rho##gh2
P1 +(1.94)(32.2)(80) = 2592 + (1.94)(32.3)(12)
P1 = -1655.8 psf
This does not make sense to me ( having a negative pressure at the top of the system)
Anyways now I will use the energy equation assume ##\alpha## = 1 also because v1 = v2 velocity head will cancel out from equation
y1 + P1/##\gamma## +##\alpha##v12/2g = y2 + P2/##\gamma## +##\alpha##v22/2g + hL
80 + (-1655.8/62.4) = 12 + (2592/62.4) + hL
hL = -.07 ft
This does not make sense to me (negative head) also I am certain that I am approaching this the wrong way because I did not use all the givens in the problem like Q, and L of pipe, and diameter of the pipe.
 
  • #6
haruspex
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I start off with Bernoulli's equation
Ok, I misunderstood what you meant by that in post #1.
Bernoulli's equation is an energy equation, but it assumes no losses. I took your "relevant equation" to be a modified Bernoulli that allowed for losses. However, I am not familiar with this form, so please explain what α and γ represent, or post a link.
I would take the inlet pressure to be atmospheric and the outlet pressure to be 18 psi higher. I am not sure what the height info means, but it sounds like there is a drop of 68 ft along the pipe. That's a bit over two atmospheres, so with no flow one would expect a pressure difference of over 30 psi.
It is not clear what form of answer is required... is it the percentage loss or rate of loss?
 
  • #7
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Divide Bernoulli's equation by ##\gamma## and include head losses. Everything is in terms of head (length) ##\gamma## = ##\rho##g alpha is coefficient correction
 
  • #8
haruspex
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Divide Bernoulli's equation by ##\gamma## and include head losses. Everything is in terms of head (length) ##\gamma## = ##\rho##g alpha is coefficient correction
Ok, that's about what I guessed. Does the "coefficient correction" represent the the loss due to drag? Or does hL represent energy loss?
 

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